A particle is projected from ground with velocity `40 m//s` at `60^@` from horizontal.
(a) Find the speed when velocity of the particle makes an angle of `37^@` from horizontal.
(b) Find the time for the above situation.
(C ) Find the vertical height and horizontal distance of the particle from the starting point in the above position. Take `g= 10m//s^2` .

Explanation

Let's solve the problem step by step.

### Given Data:
- Initial velocity, \( u = 40 \, \text{m/s} \)
- Angle of projection, \( \theta = 60^\circ \)
- Angle of interest, \( \phi = 37^\circ \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

### Part (a): Find the speed when the velocity of the particle makes an angle of \( 37^\circ \) from horizontal.

1. **Calculate the horizontal and vertical components of the initial velocity:**
- \( u_x = u \cos \theta = 40 \cos 60^\circ = 40 \times \frac{1}{2} = 20 \, \text{m/s} \)
- \( u_y = u \sin \theta = 40 \sin 60^\circ = 40 \times \frac{\sqrt{3}}{2} = 20\sqrt{3} \, \text{m/s} \)

2. **At the point where the angle is \( 37^\circ \):**
- The horizontal component of the velocity remains constant:
\[ V_x = u_x = 20 \, \text{m/s} \]

3. **Using the angle \( \phi = 37^\circ \):**
- The vertical component of the velocity at this point is:
\[ V_y = V \sin(37^\circ) \]
- The speed \( V \) can be expressed as:
\[ V = \frac{V_x}{\cos(37^\circ)} \]
- Since \( \cos(37^\circ) \approx 0.8 \):
\[ V = \frac{20}{0.8} = 25 \, \text{m/s} \]

### Part (b): Find the time for the above situation.

1. **Using the vertical motion equation:**
\[ V_y = u_y - g t \]
- Substitute \( V_y = V \sin(37^\circ) \):
\[ 25 \sin(37^\circ) = 20\sqrt{3} - 10t \]
- Since \( \sin(37^\circ) \approx 0.6 \):
\[ 25 \times 0.6 = 20\sqrt{3} - 10t \]
\[ 15 = 20\sqrt{3} - 10t \]
- Rearranging gives:
\[ 10t = 20\sqrt{3} - 15 \]
\[ t = \frac{20\sqrt{3} - 15}{10} \]
- Calculate \( t \):
\[ t \approx \frac{34.64 - 15}{10} = 1.964 \, \text{s} \]

### Part (c): Find the vertical height and horizontal distance of the particle from the starting point in the above position.

1. **Vertical height \( h \):**
- Use the vertical motion equation:
\[ h = u_y t - \frac{1}{2} g t^2 \]
- Substitute \( u_y = 20\sqrt{3} \) and \( t \approx 1.964 \):
\[ h = 20\sqrt{3} \times 1.964 - \frac{1}{2} \times 10 \times (1.964)^2 \]
\[ h \approx 34.64 \times 1.964 - 5 \times 3.866 \]
\[ h \approx 67.96 - 19.33 \approx 48.63 \, \text{m} \]

2. **Horizontal distance \( x \):**
- Use the horizontal motion equation:
\[ x = u_x t \]
- Substitute \( u_x = 20 \):
\[ x = 20 \times 1.964 \approx 39.28 \, \text{m} \]

### Summary of Results:
- (a) Speed when the angle is \( 37^\circ \): \( V \approx 25 \, \text{m/s} \)
- (b) Time to reach that angle: \( t \approx 1.964 \, \text{s} \)
- (c) Vertical height: \( h \approx 48.63 \, \text{m} \)
- (c) Horizontal distance: \( x \approx 39.28 \, \text{m} \)